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a) Hint: Plus it in

\(\sqrt{5^2+12^2}\), then solve. You should get an integer as an answer

b) To do this problem, you can square both sides.

\(x^2+2xy+y^2=x^2+y^2\)

Then cancel out x^2 and y ^2

\(2xy = 0\)

Then solve

\(x*y=0\).

So when x and y multiplied together equaling 0 is when it can satisfy the original equation.

Examples: (0, 0)  (10, 0)    (0, 5)

Values of x and y that DO NOT satisfy the original equation is when they DO NOT multiply to 0

Examples (1, 1) (2, 3) (6, 9)

I just graphed it, you can find it out!

x + y  =  sqrt (x^2 + y^2)      square both sides

x^2 + 2xy + y^2  =  x^2 + y^2

2xy  = 0

xy  =  0

Then.....this will be true as long as either x,y  = 0   or both = 0

Similarly

x + y  will not equal sqrt (x^2 + y^2)    whenever x and y are both non-zero